PH= log[H3O+]
10.25=log [H3O+]
[H3O+] = 10^10.25
[H3O+]= 1.778 ×10^10
A that’s the answer I’m not really 100% sure.
PH = -log|H₃O⁺|
pH = -log2.4*10⁻⁵ = -(log2.4+log10⁻⁵)=-(0.38-5)=-(-4.62)=4.62
Answer:
Option D. 30 g
Explanation:
The balanced equation for the reaction is given below:
2Na + S —> Na₂S
Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of S = 32 g/mol
Mass of S from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.
Thus, 30 g of S is needed for the reaction.